What Is the Value of \(a_{1000}\) in This Sequence?

[anemone]

Hi members of the forum,

I am unable to determine the value of \(\displaystyle a_{1000}\) in the problem as stated below because I think I failed to observe another useful pattern of the given sequence.

Could anyone please help me out with this problem? Thanks in advance.

Problem:
A sequence \(\displaystyle a_1\), \(\displaystyle a_2\), \(\displaystyle a_3,\;\cdots\) of positive integers satisfies the following properties:

\(\displaystyle a_1=1\)

\(\displaystyle a_{3n+1}=2a_n+1\)

\(\displaystyle a_{n+1}\ge a_n\)

\(\displaystyle a_{2001}=200\)

Find the value of \(\displaystyle a_{1000}\)

 

[chisigma]

Hi members of the forum,

I am unable to determine the value of \(\displaystyle a_{1000}\) in the problem as stated below because I think I failed to observe another useful pattern of the given sequence.

Could anyone please help me out with this problem? Thanks in advance.

Problem:
A sequence \(\displaystyle a_1\), \(\displaystyle a_2\), \(\displaystyle a_3,\;\cdots\) of positive integers satisfies the following properties:

\(\displaystyle a_1=1\)

\(\displaystyle a_{3n+1}=2a_n+1\)

\(\displaystyle a_{n+1}\ge a_n\)

\(\displaystyle a_{2001}=200\)

Find the value of \(\displaystyle a_{1000}\)

The subsequence obeys to the difference equation...

$\displaystyle a_{k+1} = 2\ a_{k}+1,\ a_{1}=1$ (1)

... the solution of which is...

$\displaystyle a_{i}= 2^{k}-1$ (2)

... and the index obeys to the difference equation...

$i_{k+1}= 3\ i_{k}+1,\ i_{1}=1$ (3)

... the solution of which is...

$i_{k} = \frac{1}{2} (3^{k}-1)$ (4)

Now we plot the sequence $a_{i}$ ...

$\displaystyle a_{1}=1,\ a_{4}= 3,\ a_{13}= 7,\ a_{40}= 15,\ a_{121}= 31,\ a_{364} = 63,\ a_{1093}= 127,\ ...$ (5)

Now, observing (5), all what we can say about $a_{1000}$ is [for the moment...] $63 \le a_{1000} \le 127$ ...

Kind regards

$\chi$ $\sigma$

 

[anemone]

The subsequence obeys to the difference equation...

$\displaystyle a_{k+1} = 2\ a_{k}+1,\ a_{1}=1$ (1)

... the solution of which is...

$\displaystyle a_{i}= 2^{k}-1$ (2)

... and the index obeys to the difference equation...

$i_{k+1}= 3\ i_{k}+1,\ i_{1}=1$ (3)

... the solution of which is...

$i_{k} = \frac{1}{2} (3^{k}-1)$ (4)

Now we plot the sequence $a_{i}$ ...

$\displaystyle a_{1}=1,\ a_{4}= 3,\ a_{13}= 7,\ a_{40}= 15,\ a_{121}= 31,\ a_{364} = 63,\ a_{1093}= 127,\ ...$ (5)

Now, observing (5), all what we can say about $a_{1000}$ is [for the moment...] $63 \le a_{1000} \le 127$ ...

Kind regards

$\chi$ $\sigma$

Hi chisigma, thank you so much for replying to this problem. But do you mean to say we could in the next step to determine what the integer value of \(\displaystyle a_{1000} is\)?

 

[Opalg]

Given that $a_{2001} = 200$, it follows that $a_{2002}\geqslant 200$. But $2002 = 3*667 + 1$, so $a_{2002} = 2a_{667}+1$. Therefore $2a_{667}+1\geqslant 200$, and $a_{667}\geqslant \frac12(199) = 99.5.$ But $a_{667}$ is an integer, and so $a_{667}\geqslant 100$. Repeating that chain of deductions, you see that $a_{223} \geqslant 50$, $a_{74} \geqslant 25$, $a_{25}\geqslant 12$, $a_8\geqslant7.$ But $a_{13}=7$ (see chisigma's post above), and so $a_8\leqslant7.$ Thus $a_8=7$, and in fact $a_n=7$ for each $n$ between $8$ and $13$ inclusive.

Next, $3*8+1=25$, so $a_{25} = 2a_8+1 = 15$, and in fact $a_n=15$ for each $n$ between $25$ and $40$ inclusive. Continue in that way to see that $a_n=31$ for each $n$ between $76$ and $121$, $a_n=63$ for each $n$ between $229$ and $364$, $a_n=127$ for each $n$ between $668$ and $1093$. In particular, $a_{1000} = 127.$

 

[chisigma]

Hi chisigma, thank you so much for replying to this problem. But do you mean to say we could in the next step to determine what the integer value of \(\displaystyle a_{1000} is\)?

We can extrapolate the result writing...

$\displaystyle \ln (y+1) \sim \ln 2x \frac{\ln 2}{\ln 3} \sim .6309 \ln 2x$ (1)

... and that leads to...

$a_{31} \sim 30.9$

$a_{364} \sim 62.9$

$a_{1093} \sim 126,9$

... so that we could conclude that...

$a_{1000} \sim 119.9$

Unfortunately the 'extra information' $a_{2001}= 200$ is not coherent with (1) because it should be $a_{2001} \sim 186.4$ (Thinking)...

Kind regards

$\chi$ $\sigma$